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3 years ago

L1 : (3, 5) and (2,5) L2: (2, 4) and (0, 4) A. Perpendicular B. Parallel C. Neither

Mathematics

1 answer:

Tresset [83]3 years ago

7 0

answer for this question is (b)parallel

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Banks and other financial institutions sometimes calculate simple interest based on

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<span>No simple interest, only compound interest.</span>

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3 years ago

TRIGONOMETRY , can you solve this plz:(<br>even if you know one of them,plz help:)

daser333 [38]

Answer:

See below

Step-by-step explanation:

$\frac{1}{1-cos A} - \frac{1}{1+cos A} =2 cosec A cot A \\ \\ L. H. S.= \frac{1}{1-cos A} - \frac{1}{1+cos A} \\ \\ = \frac{1+cos A - (1 - cos A)}{(1-cos A)(1+cos A)}\\ \\ = \frac{1+cos A - 1 + cos A}{1-cos^{2} A}\\ \\ = \frac{2cos A }{sin^{2} A}\\ \\ =2 .\frac{1}{sin A}.\frac{cos A }{sin A}.\\ \\ = 2.cosec \: A.cot \: A \\ \\ = R.H.S$

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3 years ago

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2 years ago

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F(x) = 2x + 3g(x) = 2x² + 6x + 13Find: (gºf)(x)

Butoxors [25]

Answer:

The function (gof)(x) is;

$(g\circ f)(x)=8x^2+36x+49$

Explanation:

Given the functions;

$\begin{gathered} f(x)=2x+3 \\ g(x)=2x^2+6x+13 \end{gathered}$

Solving for the function;

$(g\circ f)(x)=g(f(x))$

so, we have;

$\begin{gathered} g(f(x))=2(f(x))^2+6(f(x))+13 \\ g(f(x))=2(2x+3)^2+6(2x+3)+13 \\ g(f(x))=2(4x^2+12x+9)^{}+6(2x)+6(3)+13 \\ g(f(x))=8x^2+24x+18^{}+12x+18+13 \\ g(f(x))=8x^2+24x^{}+12x+18+18+13 \\ g(f(x))=8x^2+36x+49 \end{gathered}$

Therefore, the function (gof)(x) is;

$(g\circ f)(x)=8x^2+36x+49$

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1 year ago

What is the height of 4 1/2 and 18 cubic in.

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3 years ago

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